A person swings a 0.33 kg tether ball tied to a 5.4 m rope in an approximately horizontal circle.

(a) If the maximum tension the rope can withstand before breaking is 11 N, what is the maximum angular speed of the ball?

--I figured out the rest of the problem and i just need help with this. thanks

gravity force down = .33 * 9.8 = 3.23 N

If approximately horizontal assume swing radius = 5.4 m
m v^2/r = m w^2 r = .33*w^2*5.4 = 1.78 w^2

11 = sqrt (3.23^2 + 1.78^2 w^2)
121 = 10.4 + 3.18 w^2
w^2 = 110.5/3.18 = 34.77
w = 5.9 radians/second

To find the maximum angular speed of the ball, we need to consider the tension in the rope when it is at its maximum. At this point, the centrifugal force acting on the ball is equal to the tension in the rope to keep it in circular motion.

The centrifugal force (Fc) can be expressed as:
Fc = m * ω^2 * r

Where:
m is the mass of the ball (0.33 kg)
ω (omega) is the angular speed (in radians per second)
r is the length of the rope (5.4 m)

So, we have:
Fc = T

We'll solve for ω:
m * ω^2 * r = T

Substituting the values:
0.33 kg * ω^2 * 5.4 m = 11 N

Rearranging the equation to solve for ω:
ω^2 = 11 N / (0.33 kg * 5.4 m)

ω^2 ≈ 6.037

Taking the square root of both sides:
ω ≈ √6.037 radians/second

So, the maximum angular speed of the ball is approximately 2.46 radians per second.